Question: Let $x,$ $y,$ and $z$ be positive real numbers.  Find the minimum value of
\[\frac{4z}{2x + y} + \frac{4x}{y + 2z} + \frac{y}{x + z}.\]
Solution: Let $a = 2x,$ $b = y,$ and $c = 2z.$  Then $x = \frac{a}{2},$ $y = b,$ and $z = \frac{c}{2},$ so
\begin{align*}
\frac{4z}{2x + y} + \frac{4x}{y + 2z} + \frac{y}{x + z} &= \frac{2c}{a + b} + \frac{2a}{b + c} + \frac{b}{\frac{a}{2} + \frac{c}{2}} \\
&= \frac{2c}{a + b} + \frac{2a}{b + c} + \frac{2b}{a + c} \\
&= 2 \left (\frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} \right).
\end{align*}Let
\[S = \frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b}.\]Then
\begin{align*}
S + 3 &= \frac{a}{b + c} + 1 + \frac{b}{a + c} + 1 + \frac{c}{a + b} + 1 \\
&= \frac{a + b + c}{b + c} + \frac{a + b + c}{a + c} + \frac{a + b + c}{a + b} \\
&= (a + b + c) \left (\frac{1}{b + c} + \frac{1}{a + c} + \frac{1}{a + b} \right) \\
&= \frac{1}{2} (2a + 2b + 2c) \left (\frac{1}{b + c} + \frac{1}{a + c} + \frac{1}{a + b} \right) \\
&= \frac{1}{2} [(b + c) + (a + c) + (a + b)] \left (\frac{1}{b + c} + \frac{1}{a + c} + \frac{1}{a + b} \right).
\end{align*}By Cauchy-Schwarz,
\[[(b + c) + (a + c) + (a + b)] \left (\frac{1}{b + c} + \frac{1}{a + c} + \frac{1}{a + b} \right) \ge (1 + 1 + 1)^2 = 9,\]so
\[S \ge \frac{9}{2} - 3 = \frac{3}{2},\]and
\[\frac{4z}{2x + y} + \frac{4x}{y + 2z} + \frac{y}{x + z} \ge 2S = 3.\]Equality occurs when $a = b = c,$ or $2x = y = 2z,$ so the minimum value is $\boxed{3}.$